Physics Help?

Question

A ball is thrown horizontally from the top of
a building 25.5 m high. The ball strikes the
ground at a point 111 m from the base of the
building.
The acceleration of gravity is 9.8 m/s^2.
Find the time the ball is in motion.

Find the initial velocity of the ball.

Find the x component of its velocity just before it strikes the ground.
Find the y component of its velocity just before it strikes the ground.

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A student stands at the edge of a cliff and
throws a stone horizontally over the edge with
a speed of 15.4 m/s. The cliff is 49 m above a
flat, horizontal beach.

How long after being released does the stone strike the beach below the cliff? The acceleration of gravity is 9.8m/s^2.

At impact, what is its speed?

At what angle below the horizontal does it land?

Technobuff · Accepted Answer

a) Time to drop 25.5 metres = sqrt.(2h/g) = sqrt.(51/9.8) = 2.28 secs.
b) Initial velocity = (111/2.28) = 48.68m/sec.
c)  x component = 48.68m/sec.
d) y component = sqrt.(2gh) = 22.36m/sec.

You can do this one. Proceed as above.
For speed at impact, sqrt.(x^2 + y^2) = speed, and angle = arctan (y/x) m/sec.

? · Answer

Draw a figure!

The initial speed is vo: 
and vx = vox = constant
and vy = voy - gt , voy = 0 , so vy = -gt

y = yo - gt^2/2 , yo = 25.5 m
Time to hit ground => y=0 so  0 = 25.5 - 9.8t^2 /2 => t = 2.28 s

Range = 111 m , vox * t = 111 --> vox = 111/2.28 = 48.68 m/s
From above vy = -gt = -9.8*2.28 = - 22.34 , (the minus because vy is in the neg. y-direction)

v = sqrt(vox^2+vy^2) = sqrt(48.68^2+22.34^2) = 53.56 m/s
The angle α = arctan (vy/vox) = arctan(22.34/48.68) = 24.65°

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